Comparative Safety: 12v v 24v v 48

[evm1024]

Originally Posted by ramblinrod

I disagree.

In all of your posts on this subject, you keep disregarding or ignoring, that P = E*I (without exception).

P represent energy in Watts.
E represents voltage.
I represents current.

Energy is not proportional to current (without consideration of voltage).

The proof is that we can have lots of current (at very low voltage) through an extremely low resistance, that dissipates very little energy (won't even get slightly warm).

Energy is not proportional to voltage (without consideration of current).

The proof is that we can have a high voltage (with little current) through a very high impedance, that dissipates very little energy (won't even get slightly warm).

So there was a phenomenon I presented earlier in the thread, where some people, very knowledgeable in the electrical field, who are working with fixed voltages all the time, may consider power or energy to be directly proportional to current.

This is true and only true, if the alternate variable (voltage) remains constant. As soon as we consider a voltage change, energy is not directly related to current, because in fact, it is ALWAYS equal to voltage times current.

When we start talking about a change in both, we have to consider the impact of both on the energy dissipated.

If we increase the voltage by 400% and decrease current by 400%, the power or energy dissipated is exactly the same.

When considering a specific fault resistance, if the power sources are capable of exceeding the max current demand, (such as is often the case with batteries) the higher the voltage the greater the current, the more energy dissipated and the greater the risk of damage or fire (aka DANGER).

Now lets look at the subject or arcing.

The voltage required to cause an arc can be expressed as Voltage/Gap Distance.

This is a proportional relationship, the larger the gap the higher the voltage required to cause the arc.

Obviously, all switch and breaker contacts are designed to be capable of preventing arcing in the open state.

Obviously, when the switch contact is being made or broken, in it's travel, it will approach the separation where the arc can be initiated or extinguished.

If the power source has sufficient current available to initiate and sustain an arc, the arc will occur over a greater distance (and time) during this travel, the higher the voltage. This is a directly proportional relationship.

Now, due to the nature of arcing, once an arc is started, it is much easier to sustain over a greater gap distance.

So the more significant issue, is when the contact is opening.

Incidentally, there is likely a contact arc, when it is opened at any voltage we are discussing here, the question we are dealing with, is how damaging it is likely to be.

When the contact is opening, the arc is initiated the instant the contacts initially open, and is sustained until the contacts have travelled far enough apart, that the arc cannot be supported.

The distance that the arc can be supported is directly proportional to voltage.

Note that designs for lower voltage (12 Vdc) may not be suitable for higher voltage (48 Vdc) for this very reason.

The damage caused to the contacts is directly proportional to the energy dissipated in the arc.

The energy dissipated in the arc is calculated by P=E*I.

So it takes a higher voltage to initiate an arc and sustain it over a wider gap, and damaging energy dissipated, is also increased by the higher voltage assuming the maximum current for the circuit resistance is available, (as it will usually be, when switching high power loads.)

This is true for resistive loads, (e.g. water heater, electric range, lighting, etc.)

When we have an inductive load (e.g. motor), we have an additional consideration.

The transient voltage spike that can be generated when the circuit is suddenly opened.

In this case, the magnet field, produced by the energized coil (inductor) collapses and induces an EMF (call it "back", or "kickback", or whatever you wish) within that same coil.

Depending on the circuit, this EMF (voltage) can be much higher than the supply voltage.

There are a lot of things that come into play to determine just how high this voltage will be, but make absolutely no mistake, the higher this voltage, the more likely an arc between contacts will be initiated and the longer it will be sustained, as long as there is sufficient current available.

When we consider the energy stored in a inductor, we may determine it by considering the current through it and its impedance (resistance and inductance),but make no mistake whatsoever, the amount of current through it, everything else being equal is proportional to voltage.

Therefore the energy stored within an inductor is proportional to circuit voltage.

As we can clearly see, the relatively simple and general statement "Danger Increases with Voltage", holds true, regardless what complex situations we wish to consider, assuming there are no mitigating factors (such as something capable of limiting current, so that the energy available is reduced), and this is true and always true, regardless who claims it and what their qualifications are.

In other words, you guys with the parchments (supposedly) should know better, and the guy without, should not have to be teaching or reminding you of these principles, you have not been properly considering.

There are so many errors, misunderstandings, misinterpretations, faults in logic etc. in this post. I'm not sure where to begin.

I see that you treat voltage and current as equals, different aspects of the same thing, who knows what.

Food for thought - I can hand you a pound of current carriers (electrons in this case but I might want to neutralize their charge with a proton for safety) but I cannot hand you a pound of voltage. Electrons have mass, voltage does not. Electrons are a thing, voltage is not in the sense that you can touch an electron. Voltage is a property of an electric field and is a scalar.

Originally Posted by rr

The proof is that we can have a high voltage (with little current) through a very high impedance, that dissipates very little energy (won't even get slightly warm).

Backwards. Let me restate this.

We get a high voltage as a result of forcing very little current through a high impedance.

When did we start discussion arcing caused by the dielectric breakdown of air?

OK I give up, I'll let others who have more patience than I do take this on.

[transmitterdan]

Rod,

You seem to have little or no knowledge of electro-magnetic theories. You insist that energy cannot exist without voltage and current. This is completely incorrect and I can prove it to anyone.

If a fully charged battery is sitting on the floor disconnected from everything is there any energy stored in that battery? The obviously correct answer is yes. What power is it delivering? None. In fact, we can know accurately how much energy is in that battery. It will be stamped with an AH rating that defines the amount of energy stored therein.

But there is no current flowing in that battery. And yet there is energy stored.

The exact same thing happens in a capacitor. The amount of energy stored in any capacitor is 1/2 X C X V^2. C is the capacitance in Farads. V is the voltage. Even though no current is flowing there is energy stored therein.

An inductor stores energy in current. The amount of energy is 1/2 X L X I^2. If the resistance of the inductor is zero and you short it out the current will flow unchanged forever. Zero resistance is not possible in reality. However, superconductors do exist with such low resistance that for practical purposes it is zero. Energy can be stored in such inductors for weeks with absolutely zero voltage. Just like a battery.

Unless you can wrap your mind around these fundamental concepts we cannot have a conversation and you will continue believing things that simply are not correct.

Power = E X I as you correctly state. But power does not equal energy. Power is energy delivered per unit time. Energy becomes power when it does work such as moving a motor, making heat or creating a plasma spark. It takes energy to make a spark. That energy comes from an inductance and the current therein. Voltage has nothing to do with the problem.

[Dockhead]

Originally Posted by ramblinrod

I am missing no point; I am just politely disagreeing with the "Incorrect" ones posted based on other opinions, that are in disagreement with my own.

Obviously this is some kind of a joke -- just not quite sure what it is. But notwithstanding this assertion -- if it was indeed serious -- the point was completely missed. You were arguing about whether at all to use internal combustion. Not at all the topic.

Originally Posted by ramblinrod

This is your opinion. My opinion is that your opinion is "Incorrect!".

It is far more efficient to run a properly sized generator for longer duty cycles (even constant), than a properly sized generator of shorter duty cycles.

Where do you get this stuff? How can 24 hours of generator run time ever be more efficient than 2 hours of generator run time? 24 hours of run time means 24 hours of amortization and 24 hours of maintenance cost. If I had been using my generator like that last summer over two months, I would have been changing the oil every 4 days and I would have put on 1500 hours, or perhaps 10% of the entire useful life of the $20,000 generator, in just two months.

Originally Posted by ramblinrod

Again this is not a battery technology comparison thread. Whether on voltage is safer than another has nothing to do with battery chemistry.

And who said it does? uzzled:

Originally Posted by ramblinrod

. . . If high loads are powered by generator, there is no need for a large inverter bank.


Actually, for the argument of choosing a higher system voltage (to achieve lower current) the same argument would favour choosing to power high demand loads with the AC electrical system which is at much higher power yet, and has more safety mitigating factors (e.g. GFCIs and ELCIs) incorporated.

Something I've thought about -- big advantages of using AC power are even much greater efficiency (230v vs. 48v), more efficient motors, and simple easily available off the shelf components. Those are strong arguments.


However, the problem with that, the fatal problem, is the following: a large battery bank can easily power a short-term 10kw or even 15kW DC load, but to power an AC load you are limited by generator and inverter power. I would not want to have a 20kW generator on board, and that would likewise be an unreasonably huge inverter bank.


And in any case, I doubt that having a GFCI offsets the much greater shock danger of 230v @ 50hz (which has peak voltage of 325 volts!), compared to 48v, which is officially "Extra Low Voltage" with very little risk of shock.

[ramblinrod]

Originally Posted by evm1024

There are...

SNIP Disrespectful banter removed.

I see that you treat voltage and current as equals, different aspects of the same thing, who knows what.

Actually, I don't think you understand what I have been posting at all.

It has all been related to electrical fundamentals (Ohm's and Watt's Laws).

I'll try to explain Ohm's Law again...

E = I X R

E represents electro-motive force (EMF) in Volts

I represents current in Amps

R represents resistance in ohms

Example: If one wishes to push 10 Amps of current through 2 ohms of resistance...

E = I x R

E = 10 A x 2 ohms

E = 20 Volts

Solution: It will take 20 Volts to push 10 Amps of current through 2 ohms of resistance.

By simple mathematical manipulation, we can also derive...

I = E/R and R = E/I

(But lets not get ahead of ourselves.)

Food for thought - I can hand you a pound of current carriers (electrons in this case) but I might want to neutralize their charge with a proton for safety) but I cannot hand you a pound of voltage.

No, I'm sorry, you cannot hand me a pound of electrons.

(Some of the other random stuff you posted about electrons having mass, and voltage being an electrical property I have known since grade 6.)

We get a high voltage as a result of forcing very little current through a high impedance.

I disagree.

We require a voltage to force current through an impedance.

This is Ohm's Law.

If we have a closed circuit consisting of a voltage source, with a resistance across it, when the voltage is 0, the current is 0. When we increase the voltage, we increase the current.

Voltage is the measure of the electro-motive force it takes to produce a current through an impedance.

Using the water analogy (common for teaching basic electrical principles), electro-motive force is like water pressure, electrical current is like water flow flow, and electrical resistance is like a water restriction (e.g. a kink in a hose that resists water flow).

If we have no water pressure, we have no water flow.

Similarly, If we have no electro-motive force, we have no electrical current.

As we increase water pressure, we increase water flow.

Similarly, as we increase electro-motive force we increase electrical current.

If we increase water flow resistance (e.g. hose kink), we reduce water flow.

Similarly, as we increase electrical resistance, we reduce electrical current.

When did we start discussion arcing caused by the dielectric breakdown of air?

Quite some time ago in this thread. How did you miss it?

OK I give up...

SNIP (more disrespectful banter removed.

Please don't. I think everyone should have at least a fundamental knowledge of electrical principles.

Keep trying. I'm sure you'll get it eventually.

But this is not the best place to get a basic electrical education.

I have conducted a lot of training for large Canadian electrical and electronics manufacturing companies.

I have taught a lot of non-technical basic electrical fundamentals like Ohm's and Watts Laws.

I have also taught a lot of sales, application, and field service engineers about the new products we developed in our R&D department.

I've also introduced a lot of mechanical and electrical design engineers about a variety of basic business principles like market size, price vs volume, profit and loss, and on occasions, taught them a thing or a million about good design.

My suggestion is that you need to walk before you can run.

Get the fundamentals under your belt first, and when that is down solid, then you can consider learning some more advanced concepts.

I'm sure if you stick with it, you'll get it eventually.

[transmitterdan]

Originally Posted by ramblinrod

SNIP
We require a voltage to force current through an impedance.

You are right on this point. What you don’t seem to understand, in the case of arcing of relay or circuit breaker contacts, is the source of that voltage.

The source of the voltage is the energy that was stored in an inductor while current was flowing. The attempt to interrupt that current causes the inductor to produce a voltage high enough to cause the arc. The battery is not the source of the voltage. So it happens that the voltage causing the blue flash (ionized air) is not due to the system DC voltage but rather it is due to the amount of current in the device when the circuit opens and the device inductance. Thus the battery voltage does not determine the amount of damage caused by arcing as you believe. A 48V system is no worse than 12V and in most cases it is better when taking about contact damage due to arcs. There is less energy (not the same thing as power) to cause the arc in a 48V system of equal power compared to a 12V system. Less energy means less damage.

[fivecapes]

Mass of electron is 9.109x10^(-31)kg. It has mass and therefore on Earth it has weight.

If I'm not mistaken since we are discussing DC, the proper term is resistance, not impedance, unless there is a frequency somewhere.

[ramblinrod]

Originally Posted by Dockhead

How can 24 hours of generator run time ever be more efficient than 2 hours of generator run time?

Energy Efficiency = Power In / Power Out

If one requires 1kW/h of electrical power, it is generally far more efficient to run a 1 kW generator for 24 hours, than a 24 kW generator for one hour.

Something I've thought about -- big advantages of using AC power are even much greater efficiency (230v vs. 48v), more efficient motors, and simple easily available off the shelf components. Those are strong arguments.

I think you may be misusing the term efficient. In an of itself one voltage is no more efficient (Power in vs Power out) than another.

However, the problem with that, the fatal problem, is the following: a large battery bank can easily power a short-term 10kw or even 15kW DC load, but to power an AC load you are limited by generator and inverter power. I would not want to have a 20kW generator on board, and that would likewise be an unreasonably huge inverter bank.

Sorry, but you can't have it both ways.

If you have a daily consumption from a battery bank of 20 kWh, and you want to replenish that with 1 hour of generator run time, you must have a 20 kW generator.

Just to be clear, a 15kW load at 48 Vdc is 312 A. As you mentioned before, "Yikes!"

You would need a 350 A fuse or breaker. Any short circuit that draws less than 350 A, will not trip the over current protection. That is a whole lot of power (up to 16.8 kW) than can be smoldering away (or vapourizing in an instant) in your "backbone" somewhere.

This is why I don't recommend another high voltage system in the vessel. Most already have enough trouble not starting fires with the higher voltage AC system; the last thing they need is a higher voltage DC system.

And in any case, I doubt that having a GFCI offsets the much greater shock danger of 230v @ 50hz (which has peak voltage of 325 volts!), compared to 48v, which is officially "Extra Low Voltage" with very little risk of shock.

Yes it does.

An ELCI, pretty much eliminates the risk of electric shock or electrocution from the protected AC circuit (whatever the voltage).

There is no such device for DC.

The chance of getting an electrical shock from 48 Vdc in the wet marine environment is quite high. Electrocution risk is present but very low, but consider the severity of the consequence (pushing up daisies).

This is why I suggest, if considering 48 Vdc, consider moving the higher loads to the AC system and powering with a generator instead, and staying with a lower voltage DC system for the safety equipment, nav lights, cabin lighting, instruments, etc.

[transmitterdan]

Originally Posted by ramblinrod

Energy Efficiency = Power In / Power Out


Snip



This is why I suggest, if considering 48 Vdc, consider moving the higher loads to the AC system and powering with a generator instead, and staying with a lower voltage DC system for the safety equipment, nav lights, cabin lighting, instruments, etc.

Actually, efficiency in % = power out / power in X 100.

The last opinion isn’t supported by engineering facts or insurance loss reports.

[evm1024]

I will have to concede that I will not be able to hand Rod a pound of electrons. I will have to drive them up in a truck.

As I said due to the nature of free electrons I will bind those electrons to a proton. So let's do the math.

An electron masses in at 9.109^-31 kg and converting over to pounds requires we divide by roughly 2.2 so 1 electron masses 4.1406 ^-31 pounds.

We need to take the reciprocal to get the number of electrons in 1 pound.

1 / 4.1406^-31 = 2.415^30 electrons per pound of electrons.

There are a lot of ways to go from here but I think I'll use mols. So we divide by Avagadro's number to get the number of mols of electrons in 1 pound of electrons. 2.415^30 / 6.022^23 = 4.0105^6 mols of electrons in 1 pound of electrons.

So if I bind those electrons to some protons I end up with free hydrogen atoms. The electrons and protons are in a 1 to 1 ratio. I have 4.0105^6 moles of electrons so I know that I have 4.0105^6 mols of hydrogen. Of course free hydrogen does not stay free for very long so I end up with molecular hydrogen (H2).

Now because we have H2 we need to divide by 2 to get the moles of H2. So 4.0105^6 /2 = 2.005^6 mols of H2.

The mol weight of H2 is 2.01588 g/mol thus 2.005^6 * 2.01588 /1000 = 4041.83 kg of H2 and we convert to pounds so 4041.83 * 2.2 = 8,892 pounds of molecular hydrogen contains 1 pound of electrons.

There you go Rod 1 pound of electrons.

This ends up taking 57 cubic meters of liquid hydrogen so I better deliver it to you in more than one truck. I have not proof read the math but that does not matter. You get the idea and it was fun doing the calculations.

[evm1024]

Ohms law and does the sun rise or does the earth turn....

I posted:

Originally Posted by evm1024

We get a high voltage as a result of forcing very little current through a high impedance.

To which Rod replied:

Originally Posted by RR

I disagree.

We require a voltage to force current through an impedance.

This is Ohm's Law.

And then there was a bunch of other stuff that showed that .....

OK here we go. Ohms law at work.

Let's start with the charging up of the inductor.

WE take an inductor in series with a voltage source and some resistance. The resistance is selected so that the steady state current flowing through the inductor will be 1 amp.

It will take time for the current flowing through the inductor to stabilize. This Delta T is equal to L/R where L is the inductance in henries and R is in ohms. Back EMF and such are whats at work here.

We could say that it took X voltage to force Y current through Z resistance as predicted by ohm's law. Which is true.

I prefer to look at it as at X volts the resistance Z limited the current to Y amps. Also as predicted by ohm's law.

Moving on. Remember I said We get a high voltage as a result of forcing very little current through a high impedance. SO how does that work?

Well we have 1 amp flowing through our inductor and we open the switch. This is a really good switch and opens the contacts in no time at all. Further more for this discussion the resistance through the open switch is 1 million ohms. A high impedance if you will.

Please note that it is only the current through the inductor that is important. Once we open the switch the source voltage is gone and has no bearing.

At the moment the switch opens we have 1 amp of current flowing through the inductor and through the 1 million ohm open switch resistance.

Enter ohm's law.

E = I * R so plug in the numbers.

1 amp flowing through 1,000,000 ohms equals 1,000,000 volts

1 amp is a small current relatively speaking and thus a small current flowing though a high resistance gives a high voltage.

(The earth turns)

As we can see your disagreement with my statement is in error.

This is what Transmitterdan has been trying to tell you.




I've always been told that it is better to spend time trying to understand what the other person is saying rather than trying to make them understand your point. Good advise.

[ramblinrod]

Originally Posted by evm1024

I will have to concede...

SNIP (attempted distraction omitted).

3476 more and you'll be pretty much caught up (for this thread). ;-)

[evm1024]

Originally Posted by ramblinrod

SNIP Disrespectful banter removed.

Kettle, pot

Originally Posted by RR

No, I'm sorry, you cannot hand me a pound of electrons.

What, you don't want a pound of electrons? Do you mean you do not want a pound of electrons or that you do not a pound of electrons from me?

Let's not get personal.


SNIP %< A bunch of very basic stuff removed.

Originally Posted by RR

Quite some time ago in this thread. How did you miss it?

Wait were we talking about arcs in switches, breaker etc?

SNIP (more disrespectful banter removed.

Please don't. I think everyone should have at least a fundamental knowledge of electrical principles.

Keep trying. I'm sure you'll get it eventually.

But this is not the best place to get a basic electrical education.

I have conducted a lot of training for large Canadian electrical and electronics manufacturing companies.

I have taught a lot of non-technical basic electrical fundamentals like Ohm's and Watts Laws.

I know that my understanding is limited to the basics but they have served me well through the years.

I have also taught a lot of sales, application, and field service engineers about the new products we developed in our R&D department.

I've also introduced a lot of mechanical and electrical design engineers about a variety of basic business principles like market size, price vs volume, profit and loss, and on occasions, taught them a thing or a million about good design.

My suggestion is that you need to walk before you can run.

Get the fundamentals under your belt first, and when that is down solid, then you can consider learning some more advanced concepts.

I'm sure if you stick with it, you'll get it eventually.

I'm sure that you do not consider the quote above as disrespectful but it is.

[evm1024]

Originally Posted by ramblinrod

SNIP (attempted distraction omitted).

3476 more and you'll be pretty much caught up (for this thread). ;-)

I see that rather than discussing the truth that I or anyone could hand you a pound of electrons (well drive them up in a few trucks) and could not hand you not even 1 volt you resort to insult.

[ramblinrod]

Originally Posted by transmitterdan

Actually, efficiency in % = power out / power in X 100.

Correct!

(Whoops, my slip up, trying to respond to too many errors too fast.)

Power out over power it is.

(Only times 100 if desired to represent as a percentage.)

[ramblinrod]

Originally Posted by transmitterdan

You are right on this point.

Thank you. For a moment I thought everyone had lost all reason.

What you don’t seem to understand, in the case of arcing of relay or circuit breaker contacts, is the source of that voltage.

Ummm, yes I do, and I have understood this since grade 9, "Fundamentals of Electricity" class.

Thus the battery voltage does not determine the amount of damage caused by arcing as you believe.

Yes it does. The energy stored in the inductor is proportional to the current through it. The current through it is proportional to the system voltage.

Therefore the energy stored in the inductor is proportional to system voltage.

A 48V system is no worse than 12V and in most cases it is better when taking about contact damage due to arcs. There is less energy (not the same thing as power) to cause the arc in a 48V system of equal power compared to a 12V system. Less energy means less damage.

I disagree.

I think I may have just uncovered something else that is causing everyone's confusion, and it is along the same lines as the "fixed voltage" phenomenon I presented before.

Many technical papers may discuss I^2R generated heat.

Some may be considering this to mean current generated heat.

It doesn't.

It means electrical energy generated heat.

Electrical energy is measured in Watts, and may be referred to a P.

P (energy in Watts) = E (electro-motive force in Volts) x I (current in Amps).

This is Watts Law.

Note that we get an increase in energy if we keep the voltage constant, but increase the current.

However, when we keep the current constant and increase the voltage, we all also increase the energy.

If we substitute Ohms Law into the variables of Watts Law, we can derive..

P = I^2xR

and P = E^2/R

These can be used for the same circuit under a given set of conditions to calculate P (energy).

As we can see, either an increase in I (current) or an increase in E (voltage) increases P, and if we remember Ohms Law, the if R is fixed (and it is in the case of the inductor circuit) the only way to make I increase is to increase E.

Now if we look at the inductor energy storage formula, we get

P = 1/2 L * I^2.

where P = Energy in Watts, L = inductance in Henrys, and I is current in Amps.

So some may jump to the conclusion that the energy stored is based on the current and has nothing to do with the circuit voltage.

This is incorrect.

It is the circuit voltage that is driving the current in the inductor.

Now, when the inductor circuit is opened, all kinds of interesting things happen, the inductor magnetic field collapses which generates an EMF, which can be higher than the circuit voltage. I referred to this earlier as back EMF (some may prefer a different term). There can be other things going on with respect to circuit capacitance and resistance, that will affect the magnitude and duration of the transient voltage spike.

Note that we call it a transient voltage spike.

If the contact is open wide enough, the transient voltage is not high enough to cause an arc, and there is no transient current.

But, the higher the transient voltage the greater the risk of an arc, and the greater the energy dissipated in the arc if it ignites.

EEE the higher the circuit voltage, the higher the inductor current, the stronger the magnetic field, the higher the transient voltage spike, the greater the risk of arc and arc damage.